Abstract Nonsense

Crushing one theorem at a time

Generated Ideals and the Lattice of Ideals (Pt. I)

Point of Post: In this post we discuss the notion of generated ideals from subsets of a ring as well as the fact that the ideals of a ring form a complete modular lattice.

\text{ }


Much as was the case for normal subgroups of a group it’s true that there is some lattice like structure for ideals. In this post we shall describe the notion of  generated ideals and use it to prove that the poset \left(\subseteq,\mathfrak{L}\left(R\right)\right) of ideals of R is a complete modular lattice .

\text{ }

Generated Ideals

Intuitively what we’d like to figure out is if we have some ring R and some subset A of R can we find some left, right, or two-sided ideal which contains A and is ‘as small as possible’ (i.e. minimal with respect to containment). The first step in doing this is the following observation:

\text{ }

Theorem: Let R be a ring and \left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}} be a collection of ideals in R. Then, \displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_{\beta} is an ideal of R.

Proof: We know that the intersection of subrings are subrings, and so it suffices to prove that \displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta has the two-sided absorption property. To see this let r\in R and \displaystyle s\in\bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta be arbitrary. We note then that for every \beta\in\mathcal{B} we have that s\in\mathfrak{a}_\beta and so rs\in\mathfrak{a}_\beta. Thus, rs\in\mathfrak{a}_\beta for all \beta\in\mathcal{B} and so \displaystyle rs\in\bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta. Since r was arbitrary we see that \displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta has the left absorption property. A similar idea shows that it has the right absorption property and so \displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta is an ideal as desired. \blacksquare

\text{ }

A similar proof shows that in fact the arbitrary intersection of right or left ideals in a ring is also a left or right ideal respectively.

\text{ }

With this in mind, if R is a ring and A\subseteq R we define the ideal generated by A, denoted (A), to be equal to

\text{ }

\displaystyle \left(A\right)=\bigcap_{\mathfrak{a}\in\mathfrak{L}(R)\text{ s.t. }A\subseteq\mathfrak{a}}\mathfrak{a}\quad\bold{(1)}

\text{ }

Where \mathfrak{L}(R) denotes the set of all ideals of R.  We define the notions of the generated left ideal and generated right ideal by replacing \mathfrak{L}(R) with \mathfrak{L}_L(R) and \mathfrak{L}_R(R) respectively (where these denote the set of left and right ideals of R) and denote them (A)_L and (A)_R. If A=\{a_1,\cdots,a_n\} we denote \left(A\right) as (a_1,\cdots,a_n) (similarly for (A)_L and (A)_R). An ideal J (left, right, or two-sided) is called finitely generated

\text{ }

So we have done what we’ve set out to do, namely show that any subset of a ring R is contained inside some smallest ideal. That said, it isn’t very ‘nice looking’. As is the general case of math, once we’ve sucked the usefulness out of the abstract definition (in this case, it’s clear that our abstract desire of “minimality in terms of containment” is satisfied) we move on to a more practical definition, something we can actually work with (anyone having tried to do anything practical with singular homology understands this well). To do this though we need to make the concession that we are working with unital rings–but this really isn’t all too bad.

\text{ }

Theorem: Let R be a unital ring and A\subseteq R. Then, (A)=RAR, (A)_L=RA, and (A)_R=AR where RAR,RA, and AR are as previously defined.

Proof: We prove only that (A)=RAR since the other cases are done similarly. To do this we first note that evidently RAR\subseteq (A), since by definition every sum of the form r_1a_1s_1+\cdots+r_na_ns_n where r_k,s_k\in R and a_k\in A is any ideal \mathfrak{a}\supseteq A. So, if we can show that RAR\in\mathfrak{L}(R) we’ll be done since it will be an ideal containing A (since A=1_RA1_R\subseteq RAR) and so (A)\subseteq RAR. So, to do this we let r\in R and r_1a_1s_1+\cdots+r_na_ns_n\in RAR be arbitrary, then r(r_1a_1s_1+\cdots+r_na_ns_n)=(rr_1)a_1s_1+\cdots+(rr_n)a_ns_n\in RAR and (r_1a_1s_1+\cdots+r_na_ns_n)r=r_1a_1(s_1r)+\cdots+r_na_n(s_nr)\in RAR and thus, by the arbitrariness of r and r_1a_1s_1+\cdots+r_na_ns_n we know that RAR has the left and right absorption properties, thus it remains to show that RAR\leqslant R (as abelian groups). To see this we merely note that if r_1a_1s_1+\cdots+r_na_ns_n and r'_1a'_1s'_1+\cdots+r'_ma'_ms'_m are in RAR with m\leqslant n then there difference is equal to r''_1a''_1s''_1+\cdots+r''_{n+m}a''_{n+m}s''_{n+m} where r''_k=r_k if k\leqslant n and r''_k=-r'_k if k>n, a''_k=a_k if k\leqslant n and a''_k=-a'_k for k>n, and finally s''_k=s_k for k\leqslant n and s''_k=-s'_k for k>n. But, this is clearly an element of RAR and so RAR\leqslant R. The conclusion follows from previous discussion. \blacksquare

\text{ }

Lattice of Ideals

With this notion of generated ideals in mind we can define the lattice structure on \mathfrak{L}(R). Before we begin though we recall some of the basic concepts from order theory. Firstly, recall that if (P,\leqslant) is a partially ordered set (poset) then a subset S\subseteq P is said to have a supremum if there exist some x\in P such that s\leqslant x for all s\in S and if y\in P is such that s\leqslant y for all s\in S then x\leqslant y. It’s easy to prove that if S has a supremum it’s unique, and so we may unambiguously denote it by \sup S or \bigvee S (or even s_1\vee\cdots\vee s_n if S=\{s_1,\cdots,s_n\}). Similarly, S is said to have an infimum if there exists some x\in P such that x\leqslant s for all s\in S and if y\in P is such that s\leqslant y for all s\in S then y\leqslant x. Once again, if S has an infimum it’s unique and we may denote it \inf S or \bigwedge S (or even s_1\wedge\cdots\wedge s_n if S=\{s_1,\cdots,s_n\}).

\text{ }

A poset (P,\leqslant) is called a complete lattice if \sup S and \inf S exist for all S\subseteq P. What we now claim is that given any ring R one has that \left(\mathfrak{L}(R),\subseteq\right) is a complete lattice. Indeed:

\text{ }

Theorem: Let R be a ring. Then \left(\mathfrak{L}(R),\subseteq\right) is a complete lattice with \displaystyle \sup \left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}=\left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right) and  \displaystyle \inf\left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}=\bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta.

Proof: Let \left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}} be any subset of \mathfrak{L}(R). We have by previous theorem that \displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\in\mathfrak{L}(R), it’s clearly a lower bound for \left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}} in \left(P,\leqslant\right), and evidently if E\in\mathfrak{L}(R) is such that E\subseteq \mathfrak{a}_\beta for all \beta\in\mathcal{B} then \displaystyle E\subseteq\bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta and so \displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta is the infimum for \left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}. Similarly, we have by definition that \displaystyle\left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right) contains \displaystyle \bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta and thus contains \mathfrak{a}_\beta for each. Moreover, if E\in\mathfrak{L}(R) contains \mathfrak{a}_\beta for every \beta\in\mathcal{B} then E contains \displaystyle \bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta and thus \displaystyle \left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right)\subseteq E. Thus, \displaystyle \bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta is the supremum of \left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}. \blacksquare

\text{ }

A similar statement holds for \mathfrak{L}_L(R) and \mathfrak{L}_R(R) with (\cdot)_L and (\cdot)_R  in place of (\cdot).

\text{ }

\text{ }


1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.


June 22, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , , ,


  1. […] denoted by concatenation, to be the ideal . In other words, the product of two ideals is the ideal generated by the setwise product of the ideals. But, just as the case with the sum of ideals this formidable […]

    Pingback by Semiring of Ideals « Abstract Nonsense | June 23, 2011 | Reply

  2. […] Isomorphism Theorem’ considering the following Hasse diagram of the sets in question in the poset  (in fact, it really applies to the poset but usually the theorem is stated for when […]

    Pingback by Second Ring Isomorphism Theorem « Abstract Nonsense | June 28, 2011 | Reply

  3. […] be a lattice. We call a sublattice of if is itself a lattice. We denote this by when convenient. It’s […]

    Pingback by Fourth Ring Isomorphism Theorem (Lattice Isomorphism Theorem) (Pt. I) « Abstract Nonsense | July 9, 2011 | Reply

  4. […] this in such a language. In particular, it’s easy to see that a number is prime if the generated ideal   is proper (i.e. not equal to ) and has the property that implies or . Or, taking it one step […]

    Pingback by Prime Ideals (Pt. I) « Abstract Nonsense | August 17, 2011 | Reply

  5. […] a heuristic statement with some real truth behind it, is that a ring with a particularly simple lattice structure has a particularly simple overall ring structure. Thus, to find examples of rings which are easy […]

    Pingback by Maximal Ideals (Pt. I) « Abstract Nonsense | August 31, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: